Load Balancing with Dynamic Set of Balls and Bins

TL;DR

This paper introduces a dynamic load balancing scheme for distributing balls into bins with the ability to add or remove balls and bins efficiently, minimizing movement and lookup costs while maintaining balanced loads.

Contribution

It presents a novel hashing-style scheme that achieves near-optimal load balancing with low expected movement and lookup costs in dynamic environments.

Findings

Expected to locate a ball by checking few bins in expectation.

Expected to move few balls during insertions and deletions.

Bounds are tight for small capacities and improve for larger capacities.

Abstract

In dynamic load balancing, we wish to distribute balls into bins in an environment where both balls and bins can be added and removed. We want to minimize the maximum load of any bin but we also want to minimize the number of balls and bins affected when adding or removing a ball or a bin. We want a hashing-style solution where we given the ID of a ball can find its bin efficiently. We are given a balancing parameter $c=1+\epsilon$, where $\epsilon\in (0,1)$. With $n$ and $m$ the current numbers of balls and bins, we want no bin with load above $C=\lceil c n/m\rceil$, referred to as the capacity of the bins. We present a scheme where we can locate a ball checking $1+O(\log 1/\epsilon)$ bins in expectation. When inserting or deleting a ball, we expect to move $O(1/\epsilon)$ balls, and when inserting or deleting a bin, we expect to move $O(C/\epsilon)$ balls. Previous bounds were off by a factor $1/\epsilon$. These bounds are best possible when $C=O(1)$ but for larger $C$, we can do much better: Let $f=\epsilon C$ if $C\leq \log 1/\epsilon$, $f=\epsilon\sqrt{C}\cdot \sqrt{\log(1/(\epsilon\sqrt{C}))}$ if $\log 1/\epsilon\leq C<\tfrac{1}{2\epsilon^2}$, and $C=1$ if $C\geq \tfrac{1}{2\epsilon^2}$. We show that we expect to move $O(1/f)$ balls when inserting or deleting a ball, and $O(C/f)$ balls when inserting or deleting a bin. For the bounds with larger $C$, we first have to resolve a much simpler probabilistic problem. Place $n$ balls in $m$ bins of capacity $C$, one ball at the time. Each ball picks a uniformly random non-full bin. We show that in expectation and with high probability, the fraction of non-full bins is $\Theta(f)$. Then the expected number of bins that a new ball would have to visit to find one that is not full is $\Theta(1/f)$. As it turns out, we obtain the same complexity in our more complicated scheme where both balls and bins can be added and removed.